I need to compute that for $x \in [0, 2\pi]$
$$\sum_{n=1}^\infty\frac{\sin(nx)}{n^3} = \frac{1}{12}x(x-\pi)(x-2\pi)$$
by using the uniform convergence
$$\sum_{n=1}^\infty\frac{\sin(nx)}{n} = \frac{\pi -x}{2}$$ for each $\delta > 0$ on the interval $(\delta, 2\pi - \delta)$.
I'm afraid I don't even know how to start and would be pleased if someone helped me.
And can I use this to compute $$\sum_{n=1}^\infty\frac{1}{n^6} = \frac{\pi^6}{945}$$
by making use of the fact that if f is Riemann-integrable its Fourier series converges to f and
$$\sum_{k=-\infty}^\infty |c_k|^2 = \frac{1}{2\pi}\int_{0}^{2\pi} |f(x)|^2 \, dx$$
Thanks for help!